Question: Solve for $r$, $ -\dfrac{1}{12r - 8} = \dfrac{r + 10}{15r - 10} + \dfrac{5}{3r - 2} $
Answer: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $12r - 8$ $15r - 10$ and $3r - 2$ The common denominator is $60r - 40$ To get $60r - 40$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ -\dfrac{1}{12r - 8} \times \dfrac{5}{5} = -\dfrac{5}{60r - 40} $ To get $60r - 40$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ \dfrac{r + 10}{15r - 10} \times \dfrac{4}{4} = \dfrac{4r + 40}{60r - 40} $ To get $60r - 40$ in the denominator of the third term, multiply it by $\frac{20}{20}$ $ \dfrac{5}{3r - 2} \times \dfrac{20}{20} = \dfrac{100}{60r - 40} $ This give us: $ -\dfrac{5}{60r - 40} = \dfrac{4r + 40}{60r - 40} + \dfrac{100}{60r - 40} $ If we multiply both sides of the equation by $60r - 40$ , we get: $ -5 = 4r + 40 + 100$ $ -5 = 4r + 140$ $ -145 = 4r $ $ r = -\dfrac{145}{4}$